CIS

Geometry Difficulty 5.1 AIME, harder Prove it Soviet Union

Problem:

A plane intersects a sphere in a circle $C$ . The points $A$ and $B$ lie on the sphere on opposite sides of the plane. The line joining $A$ to the center of the sphere is normal to the plane. Another plane $p$ intersects the segment $AB$ and meets $C$ at $P$ and $Q$ . Show that $BP \cdot BQ$ is independent of the choice of $p$ .

Solution

Solution:

All points of the circle $C$ are equidistant from $A$ . The plane $p$ also meets the sphere in a circle $C'$ . Let $C''$ be the circle center $A$ radius $AP$ . Provided that $AB$ is not a diameter of $C'$ , one of the lines $BP$ , $BQ$ will meet $C''$ again at some point $R$ (see diagram).

Now since arcs $AP$ , $AQ$ are equal, so are the angles $ABP$ , $ABQ$ . Hence triangles $ABP$ , $ABQ$ are congruent and so $BP = BR$ . Hence $BP \cdot BQ = BR \cdot BQ$ . But the square of the tangent from $B$ to $C''$ is $AB^2 - AP^2$ , so $BR \cdot BQ = AB^2 - AP^2$ , which is independent of the position of $p$ .

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